Keep in mind that diode junctions also have some capacitance, and the 1N4148 has a very low leakage when reverse-biased.

So, the first time the uC's I/O pin goes low, the LED and 1N4148 junction capacitance will get charged to a certain amount. When the I/O pin goes high, the 1N4148 is now reverse-biased, and the capacitance across the LED keeps the voltage at node "a" low. The LED will eventually discharge, but it will take a very long time.

Even if you left the voltmeter in circuit for 10 days, the voltage will still be the same.

What the voltmeter measuring is not the actual voltage but +5V minus the LED drop, which is some 2.3V(LED diode drop with forward current fraction of a micro-amp).

The 1N4148 is reverse biased and only tiny tiny leakage current is flowing. This is so because voltage measurement would inevitably create a current to ground via the voltmeter, even with a 10MΩ input impedance DVM.

The only way to get +5V at that point is to use a real "potentiometer" but I bet most younger members would picture immediately of a variable resistor instead.

Keep in mind that diode junctions also have some capacitance, and the 1N4148 has a very low leakage when reverse-biased.

So, the first time the uC's I/O pin goes low, the LED and 1N4148 junction capacitance will get charged to a certain amount. When the I/O pin goes high, the 1N4148 is now reverse-biased, and the capacitance across the LED keeps the voltage at node "a" low. The LED will eventually discharge, but it will take a very long time.

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thanks for your reply. I agree that capacitance will get charged, but, i am not able to understand the meaning of ' the capacitance across the LED keeps the voltage at node "a" low.' Can you explain it detailed? thanks.

In addition, i find out a spike at node "a" when the I/O pin from low to high. Is that the result of capacitance?

Keep in mind that diode junctions also have some capacitance, and the 1N4148 has a very low leakage when reverse-biased.

So, the first time the uC's I/O pin goes low, the LED and 1N4148 junction capacitance will get charged to a certain amount. When the I/O pin goes high, the 1N4148 is now reverse-biased, and the capacitance across the LED keeps the voltage at node "a" low. The LED will eventually discharge, but it will take a very long time.

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I disagree;
Node 'b' and 'c' will be raised to 5V rather quickly.
The only issue now is node 'a', assumed to be held at 2.7V.
This gives 5-2.7 = 2.3V forward biasing the white LED.
If the white LED has this much forward bias it will charge up node 'a' to 5V.

Even if you left the voltmeter in circuit for 10 days, the voltage will still be the same.

What the voltmeter measuring is not the actual voltage but +5V minus the LED drop, which is some 2.3V(LED diode drop with forward current fraction of a micro-amp).

The 1N4148 is reverse biased and only tiny tiny leakage current is flowing. This is so because voltage measurement would inevitably create a current to ground via the voltmeter, even with a 10MΩ input impedance DVM.

The only way to get +5V at that point is to use a real "potentiometer" but I bet most younger members would picture immediately of a variable resistor instead.

What is voltage reading on the meter when you measure between "a" and +5V, using a DVM?

Click to expand...

thanks for your reply.
first, answer your question to me.
I measured voltage for nodes using a oscilloscope.
Using DC voltage range of multimeter, the voltage between "a" and +5V is 0.97V(I/O pin is high) and 2.76V(I/O pin is low).
Using DC voltage range of multimeter, the voltage of node "a" is 3.3V(I/O pin is high) and 2.1V(I/O pin is low), voltage to ground.

I disagree;
Node 'b' and 'c' will be raised to 5V rather quickly.
The only issue now is node 'a', assumed to be held at 2.7V.
This gives 5-2.7 = 2.3V forward biasing the white LED.
If the white LED has this much forward bias it will charge up node 'a' to 5V.

eblc1388's explanation I think is right on.

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Node 'a' is 2.2V as I/O pin is low.
Node 'b' increase to 5V rather quickly as I/O pin is high.
4148 reverse bias, but there is a very low leakage current.
because the capacitance of LED, and the voltage of capacitor can not change abruptly, so node 'a' will stay at 2.2V in a very short time.
then, node 'a' will get charge up by node 'b'(5V), through a leakage current. eventually, the node 'a' is up to 5V.

Using DC voltage range of multimeter, the voltage between "a" and +5V is 0.97V(I/O pin is high) and 2.76V(I/O pin is low).

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Your measurement results still show discrepancy.

You should have measured 0V using the multimeter if the MCU I/O pin is HIGH. This is apparent from both simulations done by you and me.

You must place the meter leads across the LED to get this reading, there is no other way.

If you can't get 0V, either that your I/O pin is switching, in which case measurement using multimeter is meaningless or if the I/O pin is steadily at logic HIGH, then its voltage is not +5V as you have stated in the first post. It could be at +4.03V instead.

Can you do a measurement again, but include the MCU I/O pin HIGH voltage and the +5V supply too. Please ensure that the pin level is not switching.

It is impossible to have 0.97V across the LED in between two nodes that are +5V.

I still haven't figured out why the 1N4148, I also know any voltages measured will vary depending on the input impedance of the multimeter &/or scope and also due to the tiny amount of leakage currents each component will exhibit, none of which are going to be of any sort of a constant nor very predictable value.

You should have measured 0V using the multimeter if the MCU I/O pin is HIGH. This is apparent from both simulations done by you and me.

You must place the meter leads across the LED to get this reading, there is no other way.

If you can't get 0V, either that your I/O pin is switching, in which case measurement using multimeter is meaningless or if the I/O pin is steadily at logic HIGH, then its voltage is not +5V as you have stated in the first post. It could be at +4.03V instead.

Can you do a measurement again, but include the MCU I/O pin HIGH voltage and the +5V supply too. Please ensure that the pin level is not switching.

It is impossible to have 0.97V across the LED in between two nodes that are +5V.

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According to data above, the measurement using multimeter is meaningless when the I/O pin is high. These figures are strange.
But it is correct when the I/O pin is low.

This LED circuit is part of a whole circuit. The surrounding circuit has an influence to it？maybe.

When the circuit string is forward biased, it is a low impedance circuit. It is easy to correctly measure the voltages of low impedance circuits.

When the circuit string is not conducting or reverse biased, it is a high impedance circuit (what is the effective resistance of a reverse biased diode?) Every time you connect a real meter to a new position in a high-impedance circuit, you modify the voltage because you have changed the load by the very act of attaching the meter.

Experiment:

Try hooking two 10-megohm resistors in series across 5V and then try to measure the voltage with a real meter (generally 1 to 10 megohm). It should be 2.5V, and it will be, until you attach the meter. At that point things will appear to be "wrong", but if you include the impedance of the meter in your voltage divider calculations, you will find that things will actually be "right".

Experiment 2:

Since this circuit (I/O at +5V), with the meter attached, has the 4148 reverse biased, you would find the same (or very nearly the same) voltage if you completely disconnect the wire from the 4148 and measure the voltage on the led again. At this point it should be apparent that the meter is acting as a load resistance for the LED. The actual current will not be enough to illuminate the LED (micro-amps), but it will provoke a forward voltage drop across the LED.

Experiment 3:

Parallel a resistance across your voltmeter as you measure the voltage on the circuit. This effectively reduces the impedance of your voltmeter. This will make things appear even stranger.

In general, using meters on high-impedance circuits requires that the impedance of the meter be included in calculating what results you should expect, and this changes every time you move the meter.

Experiment:
Try hooking two 10-megohm resistors in series across 5V and then try to measure the voltage with a real meter (generally 1 to 10 megohm).
Experiment 2:
Since this circuit (I/O at +5V), with the meter attached, has the 4148 reverse biased, you would find the same (or very nearly the same) voltage if you completely disconnect the wire from the 4148 and measure the voltage on the led again.
Experiment 3:
Parallel a resistance across your voltmeter as you measure the voltage on the circuit.

Click to expand...

The result of experiments is exactly same as what you said.

experiment 1:

The reading of meter is 1.66V, as i attach my multimeter. then, It can be easily to get input resistance of the multimeter, that is about 10M.

For LED lighting circuit:

experiment 2:
As i disconnected the wire from the 4148, the voltage at node 'a' is almost same as before.

experiment 3:
Paralleling a 1K resistance across my probe of scope, the voltage at node 'a' decreased from 2.7V to 2.2V, and LED light up. This verify that input resistance of probe is the load of LED.

But,there is a question about experiment 1. As i described above, the result of experiment 1 seems good. However,as i used my oscilloscope to measure voltage of same node, i found i got a strange wave on my scope.I attached the picture of wave as shown below.